Question

Given cosθ = 3/5, find the five other trigonometric function values. 15

Answer 1

$cos(\theta )=\cfrac{\stackrel{adjacent}{3}}{\underset{hypotenuse}{5}}\qquad \impliedby \textit{let's now find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \sqrt{5^2-3^2}=b\implies \sqrt{25-9}=b\implies \sqrt{16}=b\implies 4=b \\\\[-0.35em] ~\dotfill$

$sin(\theta )=\cfrac{\stackrel{opposite}{4}}{\underset{hypotenuse}{5}}\qquad \qquad tan(\theta )=\cfrac{\stackrel{opposite}{4}}{\underset{adjacent}{3}}\qquad \qquad cot=\cfrac{\stackrel{adjacent}{3}}{\underset{opposite}{4}} \\\\\\ sec(\theta )=\cfrac{\stackrel{hypotenuse}{5}}{\underset{adjacent}{3}}\qquad \qquad csc(\theta )=\cfrac{\stackrel{hypotenuse}{5}}{\underset{opposite}{4}}$